Calculating Distance for Jumping off a Slope at 45°

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SUMMARY

The discussion focuses on calculating the distance a jumper travels when launching off a slope at a 45-degree angle with an initial speed of 20 m/s and a launch angle of 15 degrees. The relevant equations for this problem include horizontal distance (x = Vox * t) and vertical distance (y = Voy * t - 1/2 * g * t^2). The user expresses difficulty in starting the problem, particularly in adapting their knowledge from a similar scenario involving a projectile launched from a building at a 30-degree angle.

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  • Knowledge of trigonometric functions for angle calculations
  • Basic grasp of gravitational acceleration (g = 9.81 m/s²)
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Homework Statement



I have to calculate disctance that body(for example a jumper) that jumps of a slope at 45 degrees with initial speed of 20 m/s and under 15 degrees.

Homework Equations


x=Vox * t
y=Voy*t - 1/2*g*t*t

y=ax + b, b=0 => y=ax
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The Attempt at a Solution



I don't know where to start. I can solve a simillar example where body has been shot from a building with a certain initial speed and under under 30°. It is simillar to this picture, but here it is difference that we shoot from a slope and we don't know at what height we only know that slope is placed under 45° with a surface.

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